Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 129334 by bramlexs22 last updated on 15/Jan/21

 x (dy/dx) = y + y^3 cos x

$$\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}\:+\:\mathrm{y}^{\mathrm{3}} \mathrm{cos}\:\mathrm{x}\: \\ $$

Answered by liberty last updated on 15/Jan/21

 let y = w^(−(1/2))  →(dy/dx) = −(1/2)w^(−3/2)  (dw/dx)   ⇒−(1/2)xw^(−3/2)  (dw/dx)−w^(−1/2) =w^(−3/2) cos x   ⇒(dw/dx)+2(w/x) = −((2cos x)/x)   take integral factor ϕ=e^(∫ (dx/x)) = x   wx = ∫−((2cos x)/x).x dx = ∫−cos x dx=−2sin x+C   y^(−2)  = ((−2sin x+C)/x) or y^2 =(x/(C−2sin x))

$$\:\mathrm{let}\:\mathrm{y}\:=\:\mathrm{w}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}^{−\mathrm{3}/\mathrm{2}} \:\frac{\mathrm{dw}}{\mathrm{dx}} \\ $$$$\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xw}^{−\mathrm{3}/\mathrm{2}} \:\frac{\mathrm{dw}}{\mathrm{dx}}−\mathrm{w}^{−\mathrm{1}/\mathrm{2}} =\mathrm{w}^{−\mathrm{3}/\mathrm{2}} \mathrm{cos}\:\mathrm{x}\: \\ $$$$\Rightarrow\frac{\mathrm{dw}}{\mathrm{dx}}+\mathrm{2}\frac{\mathrm{w}}{\mathrm{x}}\:=\:−\frac{\mathrm{2cos}\:\mathrm{x}}{\mathrm{x}} \\ $$$$\:\mathrm{take}\:\mathrm{integral}\:\mathrm{factor}\:\varphi=\mathrm{e}^{\int\:\frac{\mathrm{dx}}{\mathrm{x}}} =\:\mathrm{x} \\ $$$$\:\mathrm{wx}\:=\:\int−\frac{\mathrm{2cos}\:\mathrm{x}}{\mathrm{x}}.\mathrm{x}\:\mathrm{dx}\:=\:\int−\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}=−\mathrm{2sin}\:\mathrm{x}+\mathrm{C} \\ $$$$\:\mathrm{y}^{−\mathrm{2}} \:=\:\frac{−\mathrm{2sin}\:\mathrm{x}+\mathrm{C}}{\mathrm{x}}\:\mathrm{or}\:\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{x}}{\mathrm{C}−\mathrm{2sin}\:\mathrm{x}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com